Convergence of the series $1+a+ab+a^2b+a^2b^2+\dots+a^nb^{n-1}+a^nb^n+\dots$

Question

Consider the series $$1+a+ab+a^2b+a^2b^2+\dots+a^nb^{n-1}+a^nb^n+\dots (0<a<b)$$ 1. Use Ratio test to prove that the series converges when $b<1$ and diverges when $a>1.$

2. Will the Ratio test give any conclusion of $a<1<b$

Attempt: I got

$$\overline\lim\limits \frac{a_n}{a_{n-1}}=b$$ $$\underline\lim\limits \frac{a_n}{a_{n-1}}=a$$

series converges if $b<1$ and diverges if $a>1$

Please help for second part of the question.

Answer 1

The ratio test is inconclusive if $a < 1 < b$. To demonstrate this from the definition of the ratio test depends on the definition of the ratio test you are working with, but it must be inconclusive.

This is because we can produce values $a < 1 < b$ for which the series exhibits either behavior, i.e. just knowing that $a <1 <b$ is not enough information to know whether the series converges or diverges.

Grouping pairs of terms in the series, we can write it as

$S(a,b) = (a+1)\sum_{n=0}^{\infty} (ab)^n$

Fix any $0 < a < 1$, so $\frac{1}{a} > 1$

If we choose $1 < b < \frac{1}{a}$, then $ab < 1$ and the sum converges, but if we choose $b > \frac{1}{a} > 1$, then $ab > 1$ and the sum diverges.

Answer 2

If your definition of the ratio test matches Wikipedia, the article explicitly mentions the case where $a \lt 1 \lt b$ and says that the test is inconclusive. Check your definition of the ratio test, but I think it will agree. You can show that (assuming $a,b \gt 0$) if $ab \lt 1$ the series converges and if $ab \gt 1$ the series diverges but you weren't asked for that.